Precalculus 11

4.8 Solving Linear and Quadratic Inequalities In One Variable

Last chapter we dealt with solving for a particular point of a quadratic equation. i.e. Find zeroes and vertex of ! = #$ + # − 6 What if instead of solving for a specific point we solved for a ___________ of points.

Ø We call this solving an ___________________ Inequalities that we are going to be dealing with in this chapter have to do with either linear or quadratic functions, so we call them _____________________ and _______________________ We have for basic types of linear/quadratic inequalities:

Linear Quadratic

(# + )_______0 -#$ + )# + ._______0

(# + )_______0 -#$ + )# + ._______0

(# + )_______0 -#$ + )# + ._______0

(# + )_______0 -#$ + )# + ._______0

Instead of dealing with every ____-value, these functions now only allow certain ____-values

(specifically, those less than or greater than zero)

i.e.

A 0.5 C3 th0 27 2 6 z 2

set3 be 2y I 7 I 6

zeroes at x 3 y 5.25n 2

vertex f 0.5 5.25

qfneguality

range

linear Inequalities Quadratic Inequalities

Z ZL L

E E

y y

i q2kt I DO

aim

I

Precalculus 11

Example:

Graph the following quadratic inequality: 3# − 4 ≥ 0, then draw the possible x-values on a number line:

NOTE: The x-intercept of the equation ! = 3# − 4 is x=________. Since this is point marks the start of our solution set, we call it the _______________ value.

• If the critical points are included in our system, we write them as _______________ circles

• If the critical points are NOT included in our system, we write them as _______________ circles

Example:

Graph the following quadratic inequality: #$ − 2# − 3 < 0, then draw the possible x-values on a number line:

graph y 3N 4

Find the point where y o

Find X values where yzoat2 1 33 y o

So K Z 1.33 Is our answerIf I I

o 1.331.33

Criticalclosed s

open 0 S

l 3

I

O of i l 1 D

l 0 3

Precalculus 11

The right-hand side of the inequality doesn’t always have to equal zero. If a nonzero term exists on both sides of the equation, we can solve the problem in 2 different ways

1. Move everything to one side of the inequality and solve as before 2. Plot each side as a separate function and find the intersection rather than the x-intercept

Dividing by a _________ number will ______________ the inequality

Example:

Solve the following linear inequality: 4$ # − 1 ≤ −# + 5 Represent the solution on a number line. We can test our solutions by substituting one or more values into our inequality: On your Own:

Solve the following quadratic inequality: 5# > 2(#$ − 6) Represent the solution on a number line.

1 e 3ns znegative reverse

f z I E sets intersect at 4,1y Yz eryea value gz y

Graph both Y e Yz k E 4Find the intersection J 8Find z values that satisfy 1

O 4our inequality

Ry Lu l E sets2 iz z E 2 1 5 3 works224 i i E 2 5

O E 3

Rearrangeequation 523262 6so everything is on

Sse 7 2n 12

O 2n 12 5k1 sideO 222 Sn 12

O oPlot y 25 Sa 12 I

i s O 4Find the critical value 4 0 t.se t y

Use c V to find a solution I 4224

Precalculus 11

Example:

The length of a rectangle is 2cm greater than its width. The area of the rectangle is at least 20.($. What are the possible dimensions of the rectangle to the nearest mm? Assignment: P. 355 # 1, 3-8

EaCre

f Xtz

w x

yfifth

A 220 Bye w z zo s 3

rt 2 a 220

22 22 220

x2t2x 20 I 0 solution set 2 23,6

SI i0 3.6

Test x 4 l W 2206 4 2 20

24 2 20

So e 6 a W 4

are possible solutions